Transcript Lecture37_GeneratorsMotors

Joseph Henry (1838)

V1   N1
t
flux change
generated
by primary
so
V2 N 2

V 1 N1

V2   N 2
t
is seen
within
secondary
N2
V2 
V1
N1
If N2>N1 “step up” transformer
If N2<N1 “step down” transformer
Transformers
Power = IV
CONSERVATION OF ENERGY
tells us that for any transformer
power in = power out
P (in primary) = P (in secondary)
I1V1 = I2V2
A
B
6V
Compared to the voltage drop across
coil A, the voltage across coil B is
1) less than 6 V
2) 6 V
3) greater than 6 V
120 Watt
120 V
AC
240 V
120 V
What is the voltage
across the light bulb?
1) 60 V
2) 120 V
3) 240 V
What is the current
through the light bulb?
1) 1/2 Amp
2) 1 Amp
3) 2 Amps What is the current
through middle circuit?
1)1/2 Amp
2)1 Amp
3)2 Amps
Electric Power
Transmission
2,000 V
750,000 V
7,000 V
120 V
0.84 A
0.00224 A
0.24 A
14 A
step down transformer
high V in
low I in
low V out
high I out
energy in = energy out
step up transformer
low V in
high I in
high V out
low I out
energy in = energy out
AC power delivered to homes because
transformers do not work with DC!
b
As this copper loop rotates (counterclockwise) it turns its square face down,
away from the N-pole at left. Segment b
(highlighted in orange) rides down.
The external magnetic flux (number of
field lines) passing through the open loop
A. is increasing at this moment.
B. is decreasing at this moment.
C. remains constant.
b
As this copper loop rotates (counterclockwise) it turns its square face down,
away from the N-pole at left. Segment b
(highlighted in orange) rides down.
To resist this decrease in magnetic flux,
current is established, flowing which
way through the segment b?
A. Out of the screen, toward you.
B. Into the screen, away from you.
C. Current cannot flow through b.
B-field
B-field
As current flows through this rectangular loop
(by black s) the loop (viewed from above)
A) rotates clockwise.
B) rotates counterclockwise.
C) remains stationary.
The Lorentz Force:
I
I
I
As current flows through this rectangular loop
the back segment of the loop experiences
A) a force into the screen. C) a force up.
B) a force out of the screen. D) a force down.
E) no Lorentz force at all.
As current flows through this rectangular loop
the top segment of the loop experiences
A) a force into the screen. C) a force up.
B) a force out of the screen. D) a force down.
E) no Lorentz force at all.
As current flows through this rectangular loop
the bottom segments of the loop experiences
A) a force into the screen. C) a force up.
B) a force out of the screen. D) a force down.
E) no Lorentz force at all.
As current flows through this rectangular loop
the front segment of the loop experiences
A) a force into the screen. C) a force up.
B) a force out of the screen. D) a force down.
E) no Lorentz force at all.
electromagnets
can replace the
permanent ones
“armature”
can have
many coils
commutator
QUESTION 1
1) less than 6 V
A lot less! DC currents (like provided by this battery) cannot induce currents
at all. Only CHANGING magnetic fields (like those due to changing currents
in a solenoid) induce any currents at all! A constant input to a transformer does
NOTHING!
The 1st “step-up” transformer (4/2 =2)
doubles the voltage to 240 V, but is
QUESTION 2
followed by a “step-down” transformer that halves it.
2) 120 V
QUESTION 3
Since P =IV the 120 Watt bulb
2) 1 Amp receiving
120V, draws I=P/V=1A.
QUESTION 4
Since P=IV remains constant
Amp at every step, the current halves
when the voltage doubles.
1) 1/2
QUESTION 5 B. is decreasing at this moment.
The square loop is turning its face away from the pole.
QUESTION 6
A) Out of the screen,
toward you.
Need to replace the lost flux by generating current as shown by this right hand.
QUESTION 7
B) rotates counterclockwise.
The face of the loop turned partially toward us becomes a N-pole,
which will snap to the right, attracted to the permanent S-pole.
QUESTION 8-10
Top: C) up. Bottom: D) down.
Front: B) out.
Each from the right-hand rule of slide 13. The up and down force
might deform a loop made of weak wire, but otherwise CANCEL!
The force on the front and back segments create a torque that
help explain the motion we saw for Question 7.